Carefully selected Algebra 2 problems designed to challenge high school students and strengthen their mathematical reasoning are presented. Each problem is accompanied by a clear and detailed solution to support understanding and mastery.
Solve the quadratic equation: \[ 3x^2 - 5x - 2 = 0 \]
Use the quadratic formula: \[ x =\dfrac{-b \pm \sqrt{b^2 - 4 a c}}{2 a} = \dfrac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(-2)}}{2(3)} \]
\[ = \dfrac{5 \pm \sqrt{25 + 24}}{6} = \dfrac{5 \pm \sqrt{49}}{6} \]
\[ x = \dfrac{5 \pm 7}{6} \Rightarrow x = \dfrac{12}{6} = 2 \quad \text{or} \quad x = \dfrac{-2}{6} = -\dfrac{1}{3} \]
Write the equation of a parabola with vertex at \( (2, -3) \) and passing through the point \( (4, 5) \).
Use vertex form: \[ y = a(x - 2)^2 - 3 \]
Substitute point: \[ 5 = a(4 - 2)^2 - 3 \Rightarrow 5 = 4a - 3 \Rightarrow a = 2 \]
Equation: \[ y = 2(x - 2)^2 - 3 \]
Given that one root of the cubic polynomial \( f(x) = x^3 - 6x^2 + 11x - k \) is 2, find all the roots of \( f(x) \).
Use synthetic division to divide by \( x - 2 \):
\[ \begin{array}{r|rrrr} 2 & 1 & -6 & 11 & -k \\ & & 2 & -8 & 6 \\ \hline & 1 & -4 & 3 & 6 - k \end{array} \]
The remainder is \( 6 - k \).
For 2 to be a root, the remainder of the synthetic division \( 6 - k \) must be equal to 0: \[ 6 - k = 0 \Rightarrow k = 6 \]
Substitute \( k \) by \( 6 \) in the division above:
\[ \begin{array}{r|rrrr} 2 & 1 & -6 & 11 & - 6 \\ & & 2 & -8 & 6 \\ \hline & 1 & -4 & 3 & 0 \end{array} \]
Using the synthetic division, we write: \[ f(x) = (x - 2)(x^2 - 4x + 3) \]
Factor \( x^2 - 4x + 3 \): \[ f(x) = (x - 2)(x - 1)(x - 3) \]
So the roots of \( f(x) \) are \( 1, 2, 3 \).
Let \( f(x) = x^4 - 10x^2 + 9 \). Factor \( f(x) \) completely over the real numbers.
Let \( y = x^2 \) and hence \( y^2 = x^4 \): \[ x^4 - 10x^2 + 9 = y^2 - 10y + 9 = (y - 1)(y - 9) \]
Now substitute back: \[ f(x) = x^4 - 10x^2 + 9 = (x^2 - 1)(x^2 - 9) \]
\[ = (x - 1)(x + 1)(x - 3)(x + 3) \]
Simplify the expression: \[ \dfrac{x^2 - 9}{x^2 - x - 6} \]
Factor numerator and denominator: \[ \dfrac{x^2 - 9}{x^2 - x - 6} = \dfrac{(x - 3)(x + 3)}{(x - 3)(x + 2)} \]
Cancel the common term \( (x - 3) \): \[ \dfrac{x^2 - 9}{x^2 - x - 6} = \dfrac{x + 3}{x + 2}, \quad x \ne 3 \]
Solve the inequality: \[ \dfrac{x + 2}{x - 3} > 1 \]
Subtract \( 1 \) from both sides and simplify: \[ \dfrac{x + 2}{x - 3} - 1 > 0 \]
Write \( 1 \) as \( \dfrac{x - 3}{x - 3} \): \[ \dfrac{x + 2}{x - 3} - \dfrac{x - 3}{x - 3} > 0 \]
Subtract the fractional expressions as they have a common denominator: \[ \dfrac{x + 2 - (x - 3)}{x - 3} > 0 \]
\[ \dfrac{5}{x - 3} > 0 \]
This inequality is satisfied for: \[ x - 3 > 0 \Rightarrow x > 3 \]
Using interval notation, the solution set is written as: \[ (3 , + \infty) \]
Let \( f(x) = \dfrac{x^2 - 4x + 3}{x^2 - 5x + 6} \). Determine all vertical and horizontal asymptotes, and describe the end behavior.
Factor both: \[ f(x) = \dfrac{(x - 1)(x - 3)}{(x - 2)(x - 3)} \Rightarrow f(x) = \dfrac{x - 1}{x - 2}, \quad x \ne 3 \]
There is a Hole at: \[ x = 3 \]
Vertical asymptote: Set denominator equal to zero. \[ x - 2 = 0 \Rightarrow x = 2 \]
Horizontal asymptote: degrees of the numerator and the denominator are equal, the horizontal asymptote is given by the ratio of the leading coefficient in the numerator and denominator.
\[ \text{ratio of leading coefficient: } \dfrac{1}{1} = 1 \Rightarrow y = 1 \]
End behavior: As \( x \to \pm\infty \), \( f(x) \to 1 \) meaning that the graph stays close to the horizontal asymptote.
Solve for all real values of \( x \): \[ \left| \dfrac{2x - 5}{x + 1} \right| = 3 \]
Domain of given equation: \( x \ne -1 \)
Split into two cases:
Case 1: \[ \dfrac{2x - 5}{x + 1} = 3 \]
Cross multiply and solve: \[ 2x - 5 = 3(x + 1) \Rightarrow 2x - 5 = 3x + 3 \Rightarrow -8 = x \Rightarrow x = -8 \]
Case 2: \[ \dfrac{2x - 5}{x + 1} = -3 \]
Cross multiply and solve: \[ 2x - 5 = -3(x + 1) \Rightarrow 2x - 5 = -3x - 3 \Rightarrow 5x = 2 \Rightarrow x = \dfrac{2}{5} \]
Both solutions fall within the domain. The given equation has two solutions: \[ \left\{ -8 , \dfrac{2}{5} \right\} \]
Find all real values of \( x \) that satisfy the equation: \[ \sqrt{2x + 3} - \sqrt{-x + 7} = 1 \]
Isolate each radical on opposite sides. Add \(\sqrt{-x + 7}\) to both sides: \[ \sqrt{2x + 3} = \sqrt{-x + 7} + 1 \]
Square both sides to eliminate the first square root:
\[ \left(\sqrt{2x + 3}\right)^2 = \left(\sqrt{-x + 7} + 1\right)^2 \]
\[ 2x + 3 = (-x + 7) + 2\sqrt{-x + 7} + 1 \]
\[ 2x + 3 = -x + 8 + 2\sqrt{-x + 7} \]
Isolate the remaining square root:
\[ 2x + 3 + x - 8 = 2\sqrt{-x + 7} \]
\[ 3x - 5 = 2\sqrt{-x + 7} \]
Square both sides again:
\[ (3x - 5)^2 = (2\sqrt{-x + 7})^2 \]
\[ 9x^2 - 30x + 25 = 4(-x + 7) \]
\[ 9x^2 - 30x + 25 = -4x + 28 \]
Bring all terms to one side:
\[ 9x^2 - 26x - 3 = 0 \]
Use the quadratic formula: \[ x = \frac{-(-26) \pm \sqrt{(-26)^2 - 4(9)(-3)}}{2(9)} = \frac{26 \pm \sqrt{784}}{18} = \frac{26 \pm 28}{18} \]
\[ x = \frac{54}{18} = 3 \quad \text{or} \quad x = \frac{-2}{18} = -\frac{1}{9} \]
Check both solutions in the original equation:
Check \(x = 3\): \[ \sqrt{2(3) + 3} - \sqrt{-3 + 7} = \sqrt{9} - \sqrt{4} = 3 - 2 = 1 \quad \checkmark \]
Check \(x = -\frac{1}{9}\): \[ \sqrt{2(-\frac{1}{9}) + 3} - \sqrt{-(-\frac{1}{9}) + 7} = \sqrt{\frac{25}{9}} - \sqrt{\frac{64}{9}} = \frac{5}{3} - \frac{8}{3} = -1 \neq 1 \quad \text{Reject} \]
The real solution to the given equation is: \[ x = 3 \]
Find the inverse \( f^{-1}(x) \) of the function: \[ f(x) = \dfrac{2x - 1}{x + 3} \]
Let \( y = \dfrac{2x - 1}{x + 3} \) and interchange \( x \) and \( y \): \[ x = \dfrac{2y - 1}{y + 3} \]
Cross multiply: \[ x(y + 3) = 2y - 1 \]
Expand the left side: \[ xy + 3x = 2y - 1 \]
Group all terms with \( y \) on one side: \[ xy - 2y = -3x - 1 \]
Factor \( y \): \[ y(x - 2) = -3x - 1 \]
Divide both sides by \( (x - 2) \) and simplify to obtain: \[ y = \dfrac{-3x - 1}{x - 2} \]
The inverse function is given by: \[ f^{-1}(x) = \dfrac{-3x - 1}{x - 2} \]
If \( f(x) = x^2 + 2x + 3 \) and \( g(x) = \sqrt{x + 4} \), find the domain of \( (g \circ f)(x) \).
By definition: \[ (g \circ f)(x) = g(f(x)) = \sqrt{f(x) + 4} \]
The term \( f(x) + 4 \) inside the square root must be non-negative, hence: \[ f(x) + 4 \geq 0 \]
Substitute \( f(x) \) by its expression: \[ x^2 + 2x + 3 + 4 \geq 0 \]
Group: \[ x^2 + 2x + 7 \geq 0 \]
The discriminant of the quadratic expression \( x^2 + 2x + 7 \) is: \[ \Delta = b^2 - 4 a c = 2^2 - 4(1)(7) = - 24 \]
Since the discriminant is negative and the leading coefficient of the quadratic expression \( a = 1 \) is positive, this quadratic expression is always positive and therefore the inequality is satisfied for all real values of \( x \), so the domain of \( (g \circ f)(x) \) is: \[ (-\infty, +\infty) \]
If \( f(x) = 3^{x+1} - 2 \) and \( f(a) = f(b) \), prove that \( a = b \). Then explain what this tells you about the function.
If \( f(a) = f(b) \), then: \[ 3^{a+1} - 2 = 3^{b+1} - 2 \Rightarrow 3^{a+1} = 3^{b+1} \]
Since exponential functions are one-to-one: \[ a + 1 = b + 1 \Rightarrow a = b \]
Conclusion: \( f(x) \) is one-to-one and has an inverse function.
Let \( f(x) = \sqrt{x + 2} \) and \( g(x) = \dfrac{1}{x - 1} \). Find the domain of \( (f \circ g)(x) \).
By definition: \[ (f \circ g)(x) = f(g(x)) \]
\[ f(g(x)) = \sqrt{\dfrac{1}{x - 1} + 2} \Rightarrow \dfrac{1}{x - 1} + 2 \geq 0 \]
\[ \Rightarrow \dfrac{1 + 2(x - 1)}{x - 1} \geq 0 \Rightarrow \dfrac{2x - 1}{x - 1} \geq 0 \]
Solve using a sign chart:
Undefined at \( x = 1 \). Critical points: \( x = \dfrac{1}{2}, x = 1 \)
For \( x < \dfrac{1}{2} \): \( \quad \dfrac{2x - 1}{x - 1} \) is positive
For \( \dfrac{1}{2} < x < 1 \): \( \quad \dfrac{2x - 1}{x - 1} \) is negative
For \( x > 1 \): \( \quad \dfrac{2x - 1}{x - 1} \) is positive
The domain of \( (f \circ g)(x) \) is given by: \[ (-\infty , \dfrac{1}{2} ] \cup (1, \infty) \]
Given \( \log_2(x + 3) + \log_2(x - 1) = 3 \), solve for \( x \).
First find the domain of values of \( x \) where the solution must be. The arguments of the logarithmic expressions must be positive, hence the conditions: \[ x + 3 \gt 0 \quad \text{and} \quad x - 1 > 0 \]
The solution set to the above inequalities, which is the domain of the equation, is: \[ (1 , +\infty ) \]
Use the product rule: \[ \log_2[(x + 3)(x - 1)] = 3 \]
Transform the logarithmic equation to an exponential equation: \[ (x + 3)(x - 1) = 2^3 = 8 \]
Expand the left side and group like terms: \[ x^2 + 2x - 3 = 8 \Rightarrow x^2 + 2x - 11 = 0 \]
Use the quadratic formula to solve: \[ x = \dfrac{-2 \pm \sqrt{48}}{2} = \dfrac{-2 \pm 4\sqrt{3}}{2} = -1 \pm 2\sqrt{3} \]
Check for domain and only \[ x = - 1 + 2 \sqrt 3 \] is a solution to the given equation.
Solve for all real values of \( x \): \[ \log_3(x^2 - 4x) = \log_3(3x - 8) \]
Since logarithms have the same base and are equal, their arguments are equal: \[ x^2 - 4x = 3x - 8 \]
Rewrite the equation as: \[ x^2 - 7x + 8 = 0 \]
Use quadratic formulas: \[ x = \dfrac{7 \pm \sqrt{49 - 32}}{2} = \dfrac{7 \pm \sqrt{17}}{2} \]
Because when solving, we did not take into account the domain of the given equation, we now need to check the solutions of the given equation by calculating the left-hand side (LHS) and right-hand side (RHS) of the equation and compare them.
1) For \[ x = \dfrac{7 + \sqrt{17}}{2} \]
\[ \text{LHS} = \log_3(x^2 - 4x) = \log_3\left(\left(\dfrac{7 + \sqrt{17}}{2}\right)^2 - 4\left(\dfrac{7 + \sqrt{17}}{2}\right)\right) = \log _3\left(\dfrac{5+3\sqrt{17}}{2}\right) \]
\[ \text{RHS} = \log_3(3x - 8) = \log_3\left(3\left(\dfrac{7 + \sqrt{17}}{2}\right) - 8\right) = \log _3\left(\dfrac{5+3\sqrt{17}}{2}\right) \]
Conclusion: LHS = RHS and therefore \( x = \dfrac{7 + \sqrt{17}}{2} \) is a solution to the given equation.
2) For \[ x = \dfrac{7 - \sqrt{17}}{2} \]
\[ \text{LHS} = \log_3(x^2 - 4x) = \log_3\left(\left(\dfrac{7 - \sqrt{17}}{2}\right)^2 - 4\left(\dfrac{7 - \sqrt{17}}{2}\right)\right) \approx \log_3( -3.68465 ) \]
The LHS is not real because the argument of the log is negative. No need to check the RHS.
Conclusion: The solution of the given equation is: \[ x = \dfrac{7 + \sqrt{17}}{2} \]
Solve the system of equations: \[ \begin{cases} 2x + y = 7 \\ x^2 + y^2 = 25 \end{cases} \]
From the first equation: \[ y = 7 - 2x \]
Substitute into the second equation: \[ x^2 + (7 - 2x)^2 = 25 \Rightarrow x^2 + 49 - 28x + 4x^2 = 25 \Rightarrow 5x^2 - 28x + 24 = 0 \]
Solve with the quadratic formula: \[ x = \dfrac{28 \pm \sqrt{784 - 480}}{10} = \dfrac{28 \pm \sqrt{304}}{10} \]
\[ x = \dfrac{28 \pm 4\sqrt{19}}{10} = \dfrac{14 \pm 2\sqrt{19}}{5} \]
Find \( y \) by substituting \( x \) by its value in \( y = 7 - 2x \): \[ y = 7 - 2 \left( \dfrac{14 \pm 2\sqrt{19}}{5}\right) \]
Simplify:
\[ \begin{pmatrix}x=\dfrac{14-2\sqrt{19}}{5},\:&y=\dfrac{7+4\sqrt{19}}{5}\\ x=\dfrac{14+2\sqrt{19}}{5},\:&y=\dfrac{7-4\sqrt{19}}{5}\end{pmatrix} \]
A geometric sequence has first term \( a = 5 \) and common ratio \( r = 3 \). What is the 6th term?
\[ a_n = a \cdot r^{n - 1} = 5 \cdot 3^{5} = 5 \cdot 243 = 1215 \]
The sum of the first \( n \) terms of a geometric sequence is \( S_n = 81 \left(1 - \left(\dfrac{1}{3}\right)^n\right) \). Find the smallest \( n \) for which \( S_n > 80.99 \).
\[ S_n > 80.99 \Rightarrow 81 \left(1 - \left(\dfrac{1}{3}\right)^n\right) > 80.99 \]
\[ \Rightarrow 1 - \left(\dfrac{1}{3}\right)^n > \dfrac{80.99}{81} \]
\[ \Rightarrow - \left(\dfrac{1}{3}\right)^n > \dfrac{80.99}{81} - 1 \]
Multiply both sides of the inequality by \( -1 \) and change the symbol of inequality:
\[ \Rightarrow \left(\dfrac{1}{3}\right)^n < 1 - \dfrac{80.99}{81} \]
Take log of both sides of the inequality: \[ n \log\left(\dfrac{1}{3}\right) < \log\left(\frac{1}{8100}\right) \]
\[ \Rightarrow n > \dfrac{\log(\frac{1}{8100})}{\log(1/3)} \approx \dfrac{-3.908}{-0.4771} \approx 8.2 \]
Since \( n \) is a positive integer, its smallest value so that \( S_n > 80.99 \) is the next integer greater than \( 8.2 \): \[ n = 9 \]
Let \( A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} \), find \( A^{-1} \) if it exists.
The formula for the inverse of a 2 by 2 matrix \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is given by:
\[ A^{-1} = \dfrac{1}{\det A} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \]
Determinant: \[ \det A = (2)(4) - (-1)(3) = 8 + 3 = 11 \]
Inverse: \[ A^{-1} = \dfrac{1}{11} \begin{bmatrix} 4 & 1 \\ -3 & 2 \end{bmatrix} \]
Let \( A = \begin{bmatrix} x & 2 \\ 1 & x \end{bmatrix} \). Find all values of \( x \) for which \( A \) is not invertible.
A square matrix is non-invertible when its determinant is equal to zero:
\[ \det(A) = x^2 - 2 = 0 \]
The equation \( x^2 - 2 = 0 \) has two solutions : \( x = \sqrt 2 \) and \( x = - \sqrt 2 \)
Conclusion: Matrix \( A \) is invertible over all real numbers except \( x = \pm \sqrt 2 \).